\(n_{CaCO_3}=\dfrac{100,2}{100}=1,002\left(mol\right)\)

PTHH: Ca(OH)₂ + CO₂ ---> CaCO₃ + H₂O

                             1,002 <---- 1,002

            C₂H₅OH + 3O₂ --t^o--> 2CO₂ + 3H₂O

             0,501 <-------------------- 1,002

\(\rightarrow m_{C_2H_5OH}=0,501.46=23,046\left(g\right)\\ \rightarrow V_{C_2H_5OH}=\dfrac{23,046}{0,8}=28,8075\left(ml\right)\)

=> Độ rượu là: \(\dfrac{29,8075}{30}=96,025^o\)

C₆H₁₂O₆→ 2CO₂+ 2C₂H₅OH

0,1mo      l→            0,2mol     →m_{rượu ng/chất} =9,2g  →V_ng/chất=\(\frac{m}{D}=\frac{9,2}{0,8}=11,5ml\)

Độ rượu=40=\(\frac{Vruou\left(ngchat\right)}{Vdd}\)\(\times100\)→Vdd/thu đk= \(\frac{11,5\times100}{40}=27,85ml\)

a, \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

b, \(n_{C_2H_6O}=\dfrac{23}{46}=0,5\left(mol\right)\)

Theo PT: \(n_{O_2}=3n_{C_2H_6O}=1,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)

c, \(V_{C_2H_6O}=\dfrac{100.46}{100}=46\left(ml\right)\)

\(\Rightarrow m_{C_2H_6O}=46.0,8=36,8\left(g\right)\)

\(\Rightarrow n_{C_2H_6O}=\dfrac{36,8}{46}=0,8\left(mol\right)\)

PT: \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_2H_5ONa}=0,4\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,4.22,4=8,96\left(l\right)\)

Vr=Vhh.Đr/100=35.92/100=32.2ml

mr=D.V=0,8.32.2=25.76g

1. Theo đề ta có : nCaCO3 = nCO2 = 1,6(mol)

PTHH :

\(C2H6O+3O2-^{t0}->2CO2+3H2O\)

0,8mol........2,4mol..............1,6mol

CO2 + ca(OH)2 - > CaCO3 + H2O

a) Vkk = 5VO2 = 5.2,4.22,4 = 268,8(l)

b) Ta có : m(rượu) = 0,8.46 = 36,8(g) => V(rượu) = \(\dfrac{36,8}{0,8}=46\left(ml\right)\)

=> độ rượu = \(\dfrac{V\left(rượu-ng-chất\right)}{Vdd\left(rượu\right)}.100=\dfrac{46}{50}.100=92^0\)

PTHH :

\(C6H12O6\xrightarrow[30-33^{0C}]{men-rượu}\) 2C2H5OH + 2CO2

0,375mol..........................0,75mol......0,75mol

a) mC6H12O6(cần dùng) = 0,375.180 =67,5(g)

b) Ta có :

V(rượu ng chất ) = \(\dfrac{0,75.46}{0,8}=43,125\left(ml\right)\)

=> V(rượu 46 độ ) = \(\dfrac{43,125.100}{46}=93,75\left(ml\right)\)